Find sum of sum of all sub-sequences
Given an array of n integers. The task is to find the sum of each sub-sequence of the array.
Examples :
Input : arr[] = { 6, 8, 5 }
Output : 76
All subsequence sum are:
{ 6 }, sum = 6
{ 8 }, sum = 8
{ 5 }, sum = 5
{ 6, 8 }, sum = 14
{ 6, 5 }, sum = 11
{ 8, 5 }, sum = 13
{ 6, 8, 5 }, sum = 19
Total sum = 76.
Input : arr[] = {1, 2}
Output : 6
Method 1 (brute force):
Generate all the sub-sequence and find the sum of each sub-sequence.
Method 2 (efficient approach):
For an array of size n, we have 2^n sub-sequences (including empty) in total. Observe, in total 2n sub-sequences, each element occurs 2n-1 times.
For example, arr[] = { 5, 6, 7 }
So, the sum of all sub-sequence will be (sum of all the elements) * 2n-1.
Below is the implementation of this approach:
C++
#include<bits/stdc++.h>
using namespace std;
int sum( int arr[], int n)
{
int ans = 0;
for ( int i = 0; i < n; i++)
ans += arr[i];
return ans * pow (2, n - 1);
}
int main()
{
int arr[] = { 6, 7, 8 };
int n = sizeof (arr)/ sizeof (arr[0]);
cout << sum(arr, n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.math.*;
class GFG {
static int sum( int arr[], int n)
{
int ans = 0 ;
for ( int i = 0 ; i < n; i++)
ans += arr[i];
return ans * ( int )(Math.pow( 2 , n - 1 ));
}
public static void main(String args[])
{
int arr[]= { 6 , 7 , 8 };
int n = arr.length;
System.out.println(sum(arr, n));
}
}
|
Python3
def sm(arr , n) :
ans = 0
for i in range ( 0 , n) :
ans = ans + arr[i]
return ans * pow ( 2 , n - 1 )
arr = [ 6 , 7 , 8 ]
n = len (arr)
print (sm(arr, n))
|
C#
using System;
class GFG
{
static int sum( int []arr, int n)
{
int ans = 0;
for ( int i = 0; i < n; i++)
ans += arr[i];
return ans * ( int )(Math.Pow(2, n - 1));
}
public static void Main()
{
int []arr= { 6, 7, 8 };
int n = arr.Length;
Console.Write(sum(arr, n));
}
}
|
PHP
<?php
function sum( $arr , $n )
{
$ans = 0;
for ( $i = 0; $i < $n ; $i ++)
$ans += $arr [ $i ];
return $ans * pow(2, $n - 1);
}
$arr = array (6, 7, 8);
$n = sizeof( $arr );
echo sum( $arr , $n ) ;
?>
|
Javascript
<script>
function sum(arr, n)
{
var ans = 0;
for ( var i = 0; i < n; i++)
ans += arr[i];
return ans * Math.pow(2, n - 1);
}
var arr = [6, 7, 8];
var n = arr.length;
document.write( sum(arr, n));
</script>
|
Time complexity: O(n)
Auxiliary space: O(1)
Last Updated :
07 Jul, 2022
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